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SEQUENCES OF ANALYTIC DISKS

arXiv:math/9207202v1 [math.CV] 10 Jul 1992SEQUENCES OF ANALYTIC DISKSEvgeny A. PoletskyAbstract. The subject considered in this paper has, at least, three points of interest.Suppose that we have a sequence of one-dimensional analytic varieties in a domain inCn.

The cluster of this sequence consists from all points in the domains such that everyneighbourhood of such points intersects with infinitely many different varieties. The firstquestion is: what analytic properties does the cluster inherit from varieties?

We give asufficient criterion when the cluster contains an analytic disk, but it follows from examples ofStolzenberg and Wermer that, in general, clusters can contain no analytic disks. So we studyalgebras of continuous function on clusters, which can be approximated by holomorphicfunctions or polynomials, and show that this algebras possess some analytic properties inall but explicitly pathological and uninteresting cases.Secondly, we apply and results about clusters to polynomial hulls and maximal functions,finding remnants of analytic structures there too.And, finally, due to more and more frequent appearances of analytic disks as tools incomplex analysis, it seems to be interesting to look at their sequences to establish termi-nology, basic notation and properties.§0.

IntroductionLet us consider the following example.Let Aj be a sequence of one-dimensional irreducible analytic subvarieties of a boundeddomain D in Cn. The cluster of this sequence is the set A of all points in D such thatevery ball, centered at a point z ∈A, has non-empty intersections with infinitely manydifferent varieties Aj.

The question is: Does A remember its analytic origin? And if itdoes, what kind of analytic geometry it inherits?By the uniformization theorem for each j and for each point z ∈Aj there is a holo-morphic mapping fj of the unit disk U into D such that f(0) = z and fj(U) = Aj.1991 Mathematics Subject Classification.

Primary: 32C25; secondary: 32E20, 32F05.Key words and phrases. Sequences of analytic functions, polynomial hulls, maximal functions.Partially supported by NSF grant #DMS-9101826Typeset by AMS-TEX1

2EVGENY A. POLETSKYMoreover, all asymptotic values of fj belong to the boundary ∂D of D. It is more con-venient to consider sequences of functions fj instead of sequences of varieties Aj becausewe can use the greater power of complex analysis. Also this replacement brings into theconsideration other important cases which we are going to discuss now.Let X be a compact in Cn and ˆX is its polynomial hull.

If z is a point of ˆX then(see Th. 5.1 below) there is a sequence of holomorphic mappings fj : U →Cn such thatfj(0) = z and boundary values of fj belong to an arbitrary fixed neighbourhood V of Xeverywhere on the unit circle S except of a set of measure εj →0.

Moreover, the clusterof the sequence fj(U) belongs to ˆX. Has the cluster an analytic structure in some sense?The same question arises when one studies maximal function on the domain D in thepluripotential theory.

It can be proved that values of these function can be obtainedas solutions of variational problems for some functional Φ on the set of all holomorphicmappings of U into D. If u(z) is the extremal value of Φ then we can find the sequenceof holomorphic mappings fj of U such that fj(0) = z and Φ(fj) →u. Again, as before,boundary values of fj lie near ∂D except of the set of a small measure.

And, again, it isinteresting to know how the maximal function behaves on the cluster of fj(U).The question about the existence of analytic structure is, of course, ambiguous. Tomake it more specific we can ask whether the cluster has an analytic disk inside ornot?

This was the first question being studied, and Stolzenberg [12] and Wermer [14]provided examples of clusters which do not contain any non-trivial analytic disks. Theseare impressive examples which got titles of ”clusters without analytic structure”.

Wegive several sufficient conditions for a cluster to contain an analytic disk. But one ofmain goals of this paper is a generalization of a result of Goldmann.

In [6] Goldmanndiscovered that the Wermer’s example has the uniqueness property: if a continuousfunction can be approximated by polynomials on this cluster and is equal to zero onsome open set then it is equal to zero everywhere.So some analyticity stays in thecluster. We show in this paper that the uniqueness property holds in many quite generalcases.We also apply our results to maximal functions.

Bedford and Kalka proved in [1] thatif a such function is smooth then through every point one can draw an analytic disk suchthat the restriction of the function to this disk is harmonic. For continuous functionsthe result does not hold in this form, but if we replace analytic disks by clusters theneverything is true.

SEQUENCES OF ANALYTIC DISKS3There is one more application of introduced notions. Let {fj : U →Cn} be a uni-formly bounded sequence of holomorphic mappings with weakly converging push-forwardmeasures (fj)∗m = µ(dz, fj) of the measure m = dθ/2π on the unit circle.

The weaklimit of such measures is, evidently, a Jensen measure. Conversely, as it was provedin [3], every Jensen measure is such a limit.

We do not explore this observation hereconcentrating on geometric properties of clusters.And another goal of this paper is to develop the nomenclature for this kind of species,establish basic properties of clusters, and to show that they have quite ample geometry.I am very grateful to Norm Levenberg for useful discussions.§1. Holomorphic measuresWe shall denote by Bn(a, r) and Sn(a, r) the ball and the sphere of radius r withcenter a lying in the complex n-dimensional space Cn.

We shall omit a and r whena = 0 or r = 1 and the index n when this does not lead to a misunderstanding. We writeU for B1 and S for S1.

A standard Lebesgue measure in Cn will be denoted by m(dz).If E is a set in the closure D a domain D ⊂C then we define its harmonic measurein the following way. If E ⊂D is an open set then the harmonic measure ω(z, E, D)of the set E with respect to D is equal to the infimum of all positive superharmonicfunctions on D which are continuous and greater than 1 on E. If F is an arbitrary set inD then we define the harmonic measure ω(z, F, D) to be equal to inf ω(z, E, D), wherethe infimum is taken over all open sets E containing F. This function may be not uppersemicontinuous, so we take its regularizationlimw→z ω(z, F, D),which is superharmonic and will be denoted also by ω(z, F, D).Let f be a bounded holomorphic mapping of the unit disk into Cn.

Since f has radiallimit values almost everywhere on S, we may consider f as a measurable mapping of theclosed unit disk U and define for a Borel set E ⊂Cn a set functionµ(E, f) = 12π m(f −1(E) ∩S) = ω(0, f −1(E) ∩S, U).This is a regular Borel measure on Cn and, if f is not constant, µ(dz, f) has no atoms,i.e. the measure µ of each point in Cn is zero.

Let X be the support of µ. Evidently,

4EVGENY A. POLETSKYf(U) belongs to the polynomial hull ˆX of X, but f(U) can be less than ˆX as the exampleof Wermer in [13] shows. The point zf = f(0) is uniquely determined as a point suchthatZp(z) µ(dz, f) = p(zf)for every polynomial p on Cn.

In particular,Zz µ(dz, f) = zf.We shall call the measure µ(dz, f) by a holomorphic measure,the point zf by thecenter of the holomorphic measureµ(dz, f). An analytic disk,corresponding to thisholomorphic measure, is the image f(U) of f.Since the author could not find any references in the literature concerning such mea-sures, we shall describe some of their properties.If f = (f1, f2, .

. ., fn) then we may consider measures µ(dz, fj) on C and, evidently, fora set E, lying in the j-th coordinate complex plane, we have µ(E, fj) = µ(E × Cn−1, f).Thus, the equality µ(dz, f) = µ(dz, g) implies that µ(dz, fj) = µ(dz, gj).

The conversestatement is, in general, false.If analytic disk f(U) is given, we may consider the mapping f as a parameterizationof f(U). We have the natural method of changing parameter by composing f with amapping h : U →U.Such a reparameterization does not change the center of theanalytic disk if h(0) = 0 and does not change the holomorphic measure if and only if his inner, as the following theorem shows.Theorem 1.1.

For a holomorphic function h : U →U with h(0) = 0 measures µ(dz, f)and µ(dz, f ◦h) are equal if and only if h is an inner function, i.e. |h| = 1 a.e.

on S.Proof. If E is a Borel set in Cn then the measure µ(E, f) is equal to the measure of theset E1 = f −1(E) ∩S.

For the mapping g = f ◦h the measure µ(E, g) is equal to themeasure of the set E2 = g−1(E) ∩S = h−1(E1). Since inner functions, mapping theorigin into the origin, preserve measures on S (see [10, 19.3.4]), m(E2) = m(E1).1Conversely, for a holomorphic mapping h of U into U, we consider the set A in S,where values of radial limits of h belongs to U.

If m(A) = 0 then h is inner, so we mayassume that m(A) > 0. Let A∗be the image of A under the mapping g = f ◦h.

We1Prof. W. Rudin told me about this reference.

SEQUENCES OF ANALYTIC DISKS5denote by A1 the set f −1(A∗) ∩S and by A2 the set g−1(A∗) ∩S. Evidently, A ⊂A2and we denote by A′ the set A2 \ A′.

Thenm(A1) = m(A2) = m(A) + m(A′).Let u be the harmonic function equal to 1 on A1 and to zero on S \ A1 and let v = u ◦h.Then u(0) = v(0) = m(Ak), k = 1, 2. But the function v is equal to 0 on S \ (A ∪A′), to1 on A′ and is less then 1 on A. Therefore,m(A2) = m(A) + m(A′) > v(0) = m(A2).□Holomorphic measures are related to the pluripotential theory as the following simplereasoning shows.

A set E ∈Cn is the set of universal holomorphic measure zero if forevery nonconstant holomorphic mapping f : U →Cn the measure µ(E, f) = 0. Suchsets exist, for example, all sets, containing only one point, are of universal holomorphicmeasure zero.

In 20-th Lusin introduced the notion of P-set. A P-setE is a set inRn such that µ(E) = 0 for every finite Borel regular measure without atoms.

There arenon-trivial examples of P-sets.Theorem 1.2. (1) Every P-set in Cn is a set of universal holomorphic measure zero.

(2) Every set of universal holomorphic measure zero in Cn is pluripolar. (3) A Borel set E in the complex plane C is a set of universal holomorphic measurezero if and only if it is polar.Proof.

1) is trivial.Part 2) follows from Corollary 2.1.2 of [9].Part 3) follows from part 2) and Theorem 2.16 of [4], claiming that polar sets hasholomorphic measure zero for every non-constant bounded mapping f : U →C.□In Cn the latter statement does not hold. For example, take E = {(z, w) : ||(z, w)|| <1, w = 0}, which is pluri-polar, but µ(E, f) ̸= 0 for f(ζ) = (ζ, 0).

It is interesting todescribe all sets of universal holomorphic measure 0 in Cn. Evidently,Theorem 1.3.

Let A be a smooth complex analytic curve in a domain D ⊂Cn. A setE ⊂A is a set of universal holomorphic measure zero if and only if E is polar on A.Proof.

Follows from part 3) of Theorem 1.2□

6EVGENY A. POLETSKYOf course, we must be aware that a set E can belong to supports of two holomorphicmeasures µ(dz, f) and µ(dz, g), and µ(E, f) ̸= 0 but µ(dz, g) = 0.Example 1.1. Let E be a closed set on [−1, 1] with linear measure equal to zero butnon-polar as a set in C. Let f be a conformal mapping of U onto the upper half of Uand let g be the universal covering of U \ E. Then µ(E, f) = 0 and µ(E, g) > 0.It is interesting to find out when two holomorphic have equal holomorphic measures.It is easy to construct an example of two holomorphic mappings which measures areabsolutely continuous with respect to each other, but not equal.§2.

Jensen measures.A Jensen measure on Cnwith barycenter z0 ∈Cn is a regular Borel measure µ withcompact support such that µ(Cn) = 1 and for every plurisubharmonic function u on Cnu(z0) ≤Zu(z)µ(dz).It is clear that every holomorphic measure µ(dz, f) is a Jensen measure with barycenterzf.Let L = {fj} be a sequence of uniformly bounded holomorphic mappings of U withweakly converging holomorphic measures µ(dz, fj), i.e. for every continuous functionϕ(z) on Cn we havelimj→∞Zϕ(z) µ(dz, fj) =Zϕ(z) µ(dz)where µ is a regular Borel measure on Cn.

Then centers zfj converge to a point z0and the measure µ = µ(dz, L) is a Jensen measure with barycenter z0. The followingtheorem, claiming that every Jensen measure on Cn can be obtained as a weak limit ofholomorphic measures, was proved, basically, in [3].

Since we could not find the theoremin the form we need in [3] and for the readers’ convenience we shall give another proofof this result.Theorem 2.1. Let D be a domain in Cn and let µ be a Jensen measure with barycenterz0 ∈D and support compactly belonging to D. Then there is a sequence L = {fj} of holo-morphic mappings of U into D with holomorphic measures µ(dz, fj) weakly convergingto µ.Proof.

Let z0 = 0. Following [3] we shall prove that the set P of all weak limits ofsequences µ(dz, fj), fj : U →D, |fj| < M, and lim fj(0) = 0, is convex.

Let µ(dz, fj)

SEQUENCES OF ANALYTIC DISKS7converges to µ and µ(dz, gj) converges to ν. We shall prove that αµ + (1 −α)ν ∈P.

Wemay assume that all mappings fj and gj are holomorphic in a neighbourhood of U andfj(0) = gj(0) = 0. We consider mappingshj(ζ) = fj(ζ) + gjrjζ,defined on rings Rj = {ζ : rj ≤ζ ≤1}.

Since either |ζ| or rj/|ζ| is less than √rj on Rj,we see that for every j and every εj > 0 the set h(Rj) belongs to the εj-neighborhoodof fj(U) ∪gj(U) when rj is sufficiently small. In particular, we can find rj such thathj(Rj) ⊂D.

Let ej(ζ) be the composition of eξ and conformal equivalence of the unitdisk and the strip {ξ : ln rj < Reξ < 0}, mapping the origin into ξ = (1 −α) ln rj andthe point ζ = 1 into the origin. Evidently, mappings ej map the arc γ1 = {eiθ : −πα <θ < πα} onto the unit circle S and its complement γ2 on the circle rjS.

We considermappings pj = hj ◦ej of U into D.If ϕ is a continuous function on Cn thenZϕ(z)µ(dz, pj) = 12π2πZ0ϕ(hj(ej(eiθ)) + gj(rje−1j (eiθ))) dθ.Because of symmetry it is enough to estimate the integral12ππαZ−παϕ(fj(ej(eiθ)) + gj(rje−1j (eiθ))) dθ.Since |gj(rje−1j (eiθ))| ≤Mrj for θ ∈γ1 the previous integral is equal to12ππαZ−παϕ(fj(ej(eiθ))) dθ + δj,where the sequence of numbers {δj} converges to 0 as j →∞.If uj are harmonicfunctions on the unit disk with boundary values ϕ(fj(eiθ)) then they are uniformlybounded and, therefore, |uj(ζ) −uj(0)| < K|ζ|. Thus,uj(ej(0)) = 12π2πZ0uj(ej(eiθ)) dθ = 12ππαZ−παϕ(fj(ej(eiθ))) dθ + 12πZγ2uj(ej(eiθ)) dθ.

8EVGENY A. POLETSKYSince |uj(ej(0))−uj(0)| < Kr1−αjand |uj(ej(eiθ))−uj(0)| < Krj for θ ∈γ2, we see that12ππαZ−παϕ(fj(ej(eiθ))) dθ = αuj(0) + kj,where numbers kj converge to 0. Therefore,Zϕ(z)µ(dz, pj) →αZϕ(z) µ1(dz) + (1 −α)Zϕ(z)µ2(dz),and we proved the convexity of P.Clearly, the set P is closed.

If µ is a Jensen measure with barycenter 0 ∈D which isnot in P then there is a continuous function ϕ on Cn such thatZϕ(z) µ(dz) = a < b ≤Zϕ(z) ν(dz)for all measures ν ∈P. It was proved in [8] (see, also, [3]) that if v(z) is a maximalplurisubharmonic function on D which is less than ϕ then the infimum over all ν ∈P ofZϕ(z) ν(dz)is equal to v(0).

Since any plurisubharminic function can be approximated by continuousplurisubharmonic functions on compacts and since the support of µ is a compact in Dwe see thata =Zϕ(z) µ(dz) ≥Zv(z) µ(dz) ≥v(0) ≥b,and we get the contradiction.□§3 LeavesLet L = {fj} be an uniformly bounded sequence of holomorphic mappings of theunit disk into Cn, i.e. all sets fj(U) belong to a ball D = B(0, R) for some R > 0.We introduce the cluster cl L of L, as the set of all points z ∈Cn such that there is asequence of points ζjk ∈U with limk→∞fjk(zjk) = z.Clusters can be quite messy and pathological.

Our goal is to find inside of them some-thing nice, and to make the first move at this direction we shall use weak compactnessof holomorphic measures.

SEQUENCES OF ANALYTIC DISKS9By the Alouglou theorem [5, V.4.2], the set of measures µ(dz, fj) is compact with re-spect to the weak topology defined by integrals of continuous functions. Therefore, everysequence L = {fj} of uniformly bounded holomorphic mappings contains a subsequence{fjk} with weakly converging holomorphic measures µ(dz, fjk).

We limit ourselves tostudies of such sequences.A sequence of uniformly bounded holomorphic mappings L = {fj} is called a leaf ifmeasures µ(dz, fj) are weakly converging to the Jensen measure µ(dz) of L, which willbe denoted by µ(dz, L). Evidently, centers zfj of analytic disks fj converge to a pointzL, which we shall call the center of L.So we see that through every point of a general cluster passes a leaf made from thesame sequence.But leaves can still be quite pathological.

For example, let hj be aconformal mapping of U onto the strip Dj = {ζ = x+iy ∈C : |x| < 1, |y| < 1/j} and gjbe any sequence of holomorphic mappings of Dj (which can approximate any continuousfunction on [-1,1]). Then the sequence fj = gj ◦hj has as its cluster any cluster of asequence of continuous mappings of [-1,1], which can be extremely unpleasant.

To avoidthe consideration of such pathologies (which is not a great loss as we show later) weshould introduce a couple of definitions.A point z ∈cl L is called nonessential if there is a number r > 0 such thatlimj→∞ω(0, f −1j(Bn(z, r)), U) = 0.Other points in cl L are called essential.We shall denote the set of essential points ofL by ess L. Evidently, this set is closed.Any leaf L = {fj} determines the set supp L equal to the support of the measureµ(dz, L). Evidently supp L is compact, every point z ∈supp L is essential, and supp L ={zL} if and only if ess L = {zL}.The following theorem claims that we can get rid of non-essential points without losingtoo much.Theorem 3.1.

For every uniformly bounded sequence L = {fj} of holomorphic map-pings of U into Cn there is another sequence of such mappings M = {gj} such that:(1) cl M is the set of all essential points of cl L;(2) all points of cl M are essential;(3) if L is a leaf then M is a leaf and µ(dz, L) = µ(dz, M).

10EVGENY A. POLETSKYProof. For every nonessential point z of cl L we can find a ball Bn(z, r(z)) such thatlimj→∞ω(0, f −1j(Bn(z, 2r(z))), U) = 0.Let us choose a countable covering of the set of nonessential points by balls Bk =B(zk, rk), where zk is a nonessential point and rk = r(zk).

We denote by Cm the unionof first m closed balls Bk, k = 1, . .

., m. For every integer m there is an integer j(m)such thatω(0, f −1j(Cm), U) < 1mwhen j ≥j(m). For j(m) ≤j < j(m + 1) we consider open sets Dmj = f −1j(Cn \ Cm)in U.

Each of Dmj contains the origin and let pj be an universal holomorphic coveringmapping of Dmj by U such that pj(0) = 0.The mapping pj has radial limits a.e. on S and, by Lindel¨of theorem, gj(ζ) = fj(pj(ζ))for almost all points ζ ∈S.

If E is a subset of U then, evidently,(1)ω(0, p−1j (E), U) ≥ω(0, E, U) −1m.We define gj = fj ◦pj and let M = {gj}. Evidently, cl M is the set of all essential pointsof L.If z is an essential point of L then for every ball B(z, r), r > 0 there is a subsequence{fjk} such thatω(0, f −1jk (B(z, r)), U) > a(r) > 0.By (1)ω(0, g−1jk (B(z, r)), U) > a(r) −1m →a(r) > 0.So all points of cl M are essential.

If L is a leaf, the same reasoning shows that µ(dz, gj)weakly converge to the same measure µ(dz, L) and the sequence M = {gj} is a leaf withµ(dz, M) = µ(dz, L).□As the first application this theorem tells us that the set of essential points is quitebig.Corollary 3.1. For every leaf L the set ess L is connected.Proof.

By Theorem 3.1 ess L = cl M which is connected.□

SEQUENCES OF ANALYTIC DISKS11A sequence M = {gj} is called perfect if all points of cl M are essential. A perfectleaf M satisfying conclusions of Theorem 3.1 for some sequence L is called a perfectionof L.Let V be an open set in Cn and L = {fj} be a sequence.

For a point z ∈fj(U) wedefine the function ω(z, V, fj(U)) to be equal to the maximum of ω(ζ, f −1j(V ), U) for allζ such that fj(ζ) = z. We set ω(z, V, fj(U)) to be equal to 0 if z ̸∈fj(U).

We introducethe harmonic measure on clusters as an upper semicontinuous functionω(z, V, L) =limw→z, j→∞ω(w, V, fj(U))on cl L.If E is arbitrary set in Cn we letω(z, E, L) = inf ω(z, V, L),where the infimum is taken over all open sets V containing E.Lemma 3.1. If a point z ∈cl L is essential then for every open set E ⊂cl L, containingz, ω(zL, E, L) > 0.Proof.

Let E be an open set in cl L. We can find r > 0 such that the intersection of theball B2 = B(z, 2r) and cl L belongs to E. For an arbitrary δ > 0 there is an open set V ⊂Cn such that V ∩cl L = E and ω(zL, V, L) < ω(zL, E, L) + δ. There is an integer k suchthat fj(ζ) ∈V if fj(ζ) ∈B(z, r) and j > k. Therefore, ω(zL, V, L) > ω(zL, B(z, r), L).Since z is essential, ω(zL, B(z, r), L) > 0.□Theorem 3.2.

Let L = {fj} be a perfect leaf and X = supp L. Then the harmonicmeasure ω(z, X, L) ≡1 on cl L.Proof. Let V be an open neighbourhood of X.

We must prove that for every z ∈cl Land for every neighbourhood W of z there is a sequence of points ajk ∈U such that:(1) fjk(ajk) ∈W;(2)limk→∞ω(ajk, f −1jk (V ), U) = 1.

12EVGENY A. POLETSKYThis is evident if zL ∈W because ω(zL, V, L) = µ(Cn, L) = 1. So we may assume thatzL ̸∈W.Let Wj = f −1j(W) and let B1 ⋐W be a ball centered in z.

Since z is essential thereis a subsequence jk such that for sets Zk = f −1jk (B1) the harmonic measureω(0, Zk, U) > δ > 0.Let Vk = f −1jk (V ) ∩S and vk(ζ) = ω(ζ, Vk, U). Suppose that vk is less than ε < 1 onWjk.

If gk is an universal cover of U \ Zk such that gk(0) = 0, then radial limits of gkare in Zk on a set of measure greater than δ. Therefore, for the function uk = vk ◦gk wehaveuk(0) < (1 −δ) + εδ = 1 −δ(1 −ε) < 1.But the sequence uk(0) = vk(0) converges to 1, so there are an integer k and a pointajk ∈Wjk such thatω(ajk, f −1jk (V ), U) > ε.Since ε < 1 is arbitrary we get our theorem.□A point z ∈cl L is called totally essential if for every ball B1 = B(z, r) the lower limitof the sequence ω(0, f −1j(B1), U) is greater than 0.

Evidently, the set of totally essentialpoints is closed and all points in supp L are totally essential. A leaf is totally perfect ifall point of its cluster are totally essential.Theorem 3.3.

If L = {fj} is a perfect leaf then for every z ∈cl L there is a totallyperfect leaf M such that zM = zL, z ∈cl L, µ(dz, M) = µ(dz, L), and cl M belongs tocl L and contains all totally essential points of L.Proof. Since z is an essential point of L, it is easy to see that there is a subsequence{fjk} such that z is totally essential for this subsequence.

Its perfection contains z andthe set T of all totally essential points of L, which will continue to be totally essential.So we may assume that z is a totally essential point of L.Let n1 > 1 be the minimal positive integer such that the setE1 = {z ∈cl L : dist(z, T) ≥1n1}is non-empty. Since E1 is compact there is a finite cover of E1 by balls B1m of radius less1/2n1, centered at points of E1 and such that for each ball B1m there is a subsequence

SEQUENCES OF ANALYTIC DISKS13{fjk} withlimk→∞ω(0, f −1jk (B1m), U) = 0.We take such a subsequence {f1j} for the ball B11. For its perfection L1 the set T1 ofall totally essential points contains T and the cl L1 does not intersect B11.

If the ballB12 contains points of T1 we leave it at peace, otherwise we repeat the procedure forthe sequence {f1j}. After finite number of steps we exhaust all balls B1m and come tothe leaf Ll with the set Tl of totally essential points.

Every point of cl Ll lies within thedistance 1/n1 from points of Tl. So if n2 is the minimal positive integer such that thesetE2 = {z ∈cl Ll : dist(z, Tl) > 1n2}is non-empty, then n2 > n1.

We again take a finite cover by balls of radius less than1/2n2 and repeat the previous construction.At the end of the process we get a sequence of leaves Mk = {fkj} with perfections Lkand the sequence of balls Bk such that M1 is a subsequence of L, Mk+1 is a subsequenceMk, the ball Bk is non-essential for Mk. Sets Tk are increasing and their union is densein the intersection of clusters of Lk.

We take the sequence {gj = fjj} and let M be itsperfection.First of all, zM = zL and µ(dz, M) = µ(dz, L). Secondly, cl M belongs to the inter-section of clusters of Lk and, therefore, doesn’t contain the union of balls Bk.

And ifthe point z ∈Tk for some k then it is totally essential for {gj}. Therefore, cl M consistsfrom totally essential points only.□Let us give an example of a perfect but not a totally perfect leaf.Example 3.1.

Let fj = (ζ, ζj) be a sequence of mappings of U into U 2. It is easy tosee that measures µ(dz, fj) converge to the measureµ = 12π dα dβon S × S. Therefore, the sequence of fj determines the leaf L = {fj} with µ(dz, L) = µ.So supp L = S × S, andcl L = {(z1, z2) : |z1| < 1, z2 = 0 or |z1| = 1, |z2| ≤1}.The leaf L is totally perfect.

14EVGENY A. POLETSKYWe can consider another sequence of holomorphic mappings gj equal to fj if j is even,and to (ζj, ζ) if j is odd. Then holomorphic measures µ(dz, gj) also converge to µ, so thesequence M = {gj} is a leaf, and this leaf is perfect but not totally perfect.

Its clusteris the union of two perfect leaves.§4. Geometry and analytic structures on leavesIf L = {fj} is a leaf in Cn then it is interesting to find out what does cl L rememberabout its holomorphic origin.

There is no sense to look at leaves with trivial Jensenmeasures: their clusters can be as pathological as clusters of continuous mappings. Sowe assume that supp L ̸= {zL}.

In this case, the set ess L ̸= {zL} and, by Theorem 3.1,we can find a perfect leaf with the same support and the same Jensen measure, whichcluster is ess L. Since remaining points again can form extremely pathological set weshall concentrate on studies of perfect leaves.The first question which naturally arises is the question about the existence of analyticdisks in clusters. The negative answer to this question was given by Stolzenberg in [12]and Wermer in [14].

There are perfect leaves such that any holomorphic mapping of theunit disk into the cluster of such leaves is constant. Before we go further, we give thebrief description of these examples.Examples of Stolzenberg and Wermer.

Both examples deal with complex analytic1-dimensional irreducible varieties Vj (Stolzenberg) and Wj (Wermer) in the unit bidiskin C2.Each of Vj or Wj contains the origin.So, according to our approach, thesesequences of varieties or their subsequence are leaves with supports lying on the boundaryof U 2. It was proved by Stolzenberg and Wermer that clusters of these leaves does notcontain an analytic disk.There is a difference in this examples.

The example of Stolzenberg is a ”Swiss cheese”lifted to C2, the projections of its cluster on both coordinate planes are nowhere dense.But if one makes a ”Swiss cheese” on a plane then it may happen that its cluster (whatis left after we drag out holes) consists from non-essential points only. By Theorem 3.1Stolzenberg’s example has a connected set of essential points joining the origin with theboundary of U 2.Varieties Wj in Wermer’s example have proper projections on one of coordinate lines,say, {w = 0}.

Moreover, there is a countable dense set of points {pk} on this line suchthat for each j the variety Wj has finite number of points over each of pk, and there

SEQUENCES OF ANALYTIC DISKS15is only one point in Wj over the origin. This property was used by Goldmann in [6]to show that the cluster of Wermer’s example is perfect.

In the same paper Goldmanndiscovered that this cluster has some analytic properties, which we discuss later. Thisdiscovery gave us the starting push for our studies.

But before we move to this questionswe want to give a sufficient conditions for the existence of analytic disks in clusters.Theorem 4.1. Let L = {fj} be a leaf in Cn.Suppose that there is a holomorphicfunction h, defined on a neighbourhood V of cl V , and numbers b < 1 and α > 0 suchthat:(1) |h| and ||∇h|| are less than M < ∞on V ;(2) |h(zL)| ≤b < 1;(3) for every integer j ≥1 there is an arc γj with the length greater than α such thatlimr→1|h(fj(rζ))| ≥1for all points ζ in γj.Then there is a non-constant holomorphic mapping g = (g1, .

. ., gn) : U →cl L such thatat least one of functions gk, 1 ≤k ≤n, covers univalently a disk of a positive radiusr = r(M, b, α) in C.Proof.

Let S(c) be a set of holomorphic functions u on U with the Bloch norm||u||B = supζ∈U|u′(ζ)|(1 −|ζ|2) ≥c.By Theorems 2 and 1 from [7] functions hj = h ◦fj belong to S(c) for some c =c(M, b, α) > 0. Since ||∇h|| < M there is an integer k(j) such that||fjk(j)||B ≥cnM ,where fjk is the k-th coordinate function of fj.

So for some k between 1 and n we canfind an infinite subsequence fjl such that:(1)|f ′jlk|(1 −|ζjl|2) ≥c2nMat some point ζjl in U;(2) the sequence fjl(ζjl) converges to z0.

16EVGENY A. POLETSKYMappingsfjl ζ −ζjl1 −ζjlζ!converge to a non-constant mapping g which image lies in cl U and |g′k(0)| ≥c/2nM. So,by the Bloch theorem, gk covers univalently a disk of a positive radius r = r(M, b, α).□Examples of Stolzenberg and Wermer (see, also, [7]) show that it is hard to awaitsomething really good at this situation but in the one-dimensional case we can do a littlebit better.Theorem 4.2.

Let L = {fj} be a perfect leaf in C and ∆be the boundary of cl L. Then∆⊂supp L.Proof. Suppose that there is a disk B1 = B(z0, r) such that z0 ∈∆, r > 0 andµ(B1), L) = 0.

We can find a point z such that a closed disk B2 = B(z, s) ⊂B(z0, r/4)does not intersect cl L.Therefore, there is an integer N such that fj(U) ∩B2 = ∅when j > N. Let B3 = B(z, r/2) and let u(z) be a harmonic function on C \ B(z, s),equal to 1 on ∂B2 and to 0 on ∂B3. Functions uj(ζ) = u(fj(ζ)) will be harmonic onU.If Ej = {ζ ∈S : uj(ζ) > 0} then the sequence of aj = m(Ej) tends to 0.IfB3 = B(z0, r/4) then u(z) > c > 0 on B3.

Harmonic functions vj on U with boundaryvalues equal to 1 on Ej and to 0 on S \Ej are greater than uj. The harmonic measure ofthe set Fj = {ζ ∈U : vj(ζ) > c} is less than aj/c and, therefore, the harmonic measureof sets Gj = {ζ ∈U : uj(ζ) > c} is less than aj/c which means that the point z0 isnonessential.□In general, it may happen that µ(∆∩B(z, r), L) = 0 for some z ∈∆.Corollary 4.1.

Let L = {fj} be a leaf in C such that zL = 0 and supp L lies outsidethe disk B1 = B(0, r) for some r > 0. Then B1 belongs to cl L.Proof.

We may assume that the leaf L is perfect. Since supp L does not intersect B1,the boundary of cl L does not intersect B1.

So B1 belongs to cl L.□Example 4.1. Let fj = (gj, hj) be uniformizations of curves Vj from the Stolzenberg’sexample by analytic disks such that fj(0) = (0, 0).

We may assume that {fj} is a leafL. Then supp L ⊂∂U 2 and, therefore, either the leaf G = {gj} or the leaf H = {hj}is non-trivial.

Their perfections are also non-trivial. Nevertheless, their clusters do notcontain any disk in C. In this example, cl L = supp L = ∆.

SEQUENCES OF ANALYTIC DISKS17So the complex geometry is not inherited by leaves explicitly. To go further we in-troduce for each leaf L algebras P(L) and H(L) of functions on cl L, which can beapproximated on cl L either by polynomials or by functions holomorphic in a neighbour-hood of cl L. Such algebras are called analytic if a function from the algebra is equal to0 on cl L provided it is equal to 0 on some open set V in cl L.Our first step is to prove the two constant theorem for functions in H(L).Theorem 4.3.

Let L = {fj} be a sequence of uniformly bounded analytic disks and leth ∈H(L). If |h| is less than M on cl L and less than m < M on an a set E ∈cl L, then|h(z)| < mdM 1−d,where d = ω(z, E, L).Proof.

Let us fix some ε > 0. The function h is the limit of functions hk holomorphic inneighbourhoods Dk of cl L. We may find an integer k such that |hk| < M ′ = (1 + ε)Mon Dk, |hk| < m′ = (1 + ε)m on E, and |hk(z) −h(z)| < ε.Let V be the open set of points in Dk, where |hk| is less than m′.Evidently,ω(z, V, L) ≥d.If Vj = f −1j(V ) then, by the definition of ω(z, V, L), there is a se-quence of points ajl ∈U such that points fjl(ajl) converge to z and the limit d′ of thesequence djl = ω(ajl, Vjl, U) is greater than d. We may assume that jl = j.If vj = hk ◦fj then by the two constants theorem|vj(aj)| < (m′)dj(M ′)1−djand|hk(z)| < (m′)d′(M ′)1−d′ ≤(m′)d(M ′)1−d.Therefore, |h(z)| < mdM 1−d.□Corollary 4.2.

Let L = {fj} be a perfect leaf in Cn and let zL be the center of L. If afunction h ∈H(L) is equal to zero on an open set E ⊂cl L then h(zL) = 0.Proof. This statement follows from the previous theorem because all points in cl L areessential and, therefore, ω(zl, E, L) > 0.□Corollary 4.3.

Let L = {fj} be a perfect leaf in Cn. Then:(1) every function h ∈H(L) attains its maximum modulus value on supp L;(2) cluster of L belongs to the polynomial hull of supp L.

18EVGENY A. POLETSKYProof. Part 1) follows immediately from Theorem 4.3 and 3.2 and part 2) is the conse-quence of part 1).□The maximum modulus principle proved just now can be obtained from the factthat the measure µ(dz, L) is an Jensen measure for H(L) at zL.

We did not explorethis approach because we intended to be as geometric as possible. Connections of thissubject with the theory of uniform algebras are evident.

Nevertheless, it seems to us thatthis theory does not allow to look closely at geometric structures generated by leaves.Moreover, we shall prove in §5 that polynomial hulls can be foliated by perfect leaves, soour method can be applied be applied to uniform algebras as well. Let us consider thefollowing example.Example 4.2.

Let L be a perfect leaf from Example 3.1. The function h(z1, z2) = w isequal to 0 on a neighbourhood of origin and is not 0 everywhere.

So perfect leaves lackof the propagation of zeros. To understand this phenomenon we must look closer at thegeometry of leaves.It happens that leaves generate complex substructures in a natural way which we aregoing to describe now.

LetGa(ζ) = ζ + a1 + aζbe a conformal automorphism of the unit disk. Let L = {fj} be a perfect leaf.

For apoint z ∈cl L we consider points ak ∈U such that the a subsequence fjk(ak) convergesto z. From the sequence of mappings gk = fjk ◦Gak we can choose a subsequence hmwith weakly converging measures µ(dz, hm).

A subleaf Lz is the perfection of the leafM = {hk}.If M is a subleaf of L then cl M ⊂cl L but it may happen that supp M is not a subsetof supp L. For example, if fj are an universal holomorphic coverings of U \ B(1/2, 1/j)by U such that fj(0) = 0, then there are subleaves M = L 12 such that µ( 12, M) = a,where a is any number between 0 and 1. A subleaf Lz such that supp Lz ⊂supp L iscalled a principal subleaf.The set of all subleaves of the leaf L is partially ordered by the inclusion relation oftheir clusters.

A leaf L is called minimal at a pointz ∈cl L if it contains no propernon-trivial subleaves passing through z. Evidently, a leaf L, minimal at zL, is totallyperfect. A leaf L is called minimal if it is minimal at all points of cl L. A point z ∈cl Lis called inner if there is a non-trivial subleaf Lz.

We shall say that a sequence of leaves

SEQUENCES OF ANALYTIC DISKS19{Lj} converges to a leaf L if every neighbourhood of every point of cl L contains pointsfrom clusters of infinitely many different leaves Lj and µ(dz, Lz) = lim µ(dz, Lj)In Example 4.2 there is only one subleaf passing through the origin, so L is minimalat zL. The leaf M from Example 3.1 has two subleaves passing through the origin so itis not minimal at zL.

Both leaves have subleaves at the boundary of the bidisk so theyare not minimal. These subleaves are analytic disks foliating the boundary.

If we takethe leaf L and a point z0 = (z1, z2) such that |z1| = 1 and z2 ̸= 0, |z2| < 1, then for eachj we can find a point aj ∈U such that fj(aj) converges to z0 and let gj = fj ◦Gaj. Wemay assume that the sequence gj ia a leaf L′ = {gj}, otherwise we take a convergingsubsequence.Evidently, cl L = cl L′, but the leaf L′ is not perfect.The origin is anon-essential point of M. We shall show that the property for a leaf to be minimal andthe analyticity of the algebra H(L) are closely related.Lemma 4.1.

(1) If points zj ∈cl L converge to a point z and measures µ(dz, Lzj) weakly convergeto a measure µ then subleaves Lzj converge to a subleaf Lz;(2) if J is a totally ordered set and L = {Lzj, j ∈J} is a set of subleaves such thatcl Lzi ⊂cl Lzj for i > j, then there is a countable sequence {Lj} in L convergingto a subleaf which belongs to all subleaves in L;(3) if z belongs to the cluster of a perfect leaf L but doesn’t belong to its support thenthere is a principal subleaf Lz passing through z.Proof. 1) If leaves Lzj = {gjk = fj ◦Gajk} then we can find a subsequence am = ajmkmin U such that measures µ(dz, gjmkm) converge to the measure µ.

Then the sequencefjm ◦Gam defines a leaf Lz.2) Let G be the intersection of clusters of all leaves in L. There is a countable sequencejk such that the set G is the intersection of cl Ljk. Let us select another subsequencewith converging holomorphic measures.

This subsequence of subleaves converges to asubleaf Lz due to the first part of the lemma.3) Suppose that there is a point z ∈cl L such that dist(z, supp L) = r > 0 and forevery Lz the support of Lz doesn’t belong to supp L. This means that there are numbersa > 0 and d > 0, and a ball B1, centered at z, such thatlimj→∞µ(W, fj ◦Gζ) > a

20EVGENY A. POLETSKYfor every point w in B1 and every ζ ∈f −1(w) and for the set W of points in Cn, withdistances to supp L greater or equal to d. Otherwise, we can find a subsequence fjk = gk,a sequence wk ∈Cn converging to z, and a sequence ζk ∈g−1k (wk) such that for thecomplement V of supp L measures µ(V, gk ◦Gζk) converge to 0, and, therefore, there isa subleaf with the support lying in the support of L.Let Ej be a subset of S such that fj(ζ) ∈W for every ζ ∈Ej. If Dj = f −1j(B1)then the function ω(ζ, Ej, U) is greater than a on Dj when j is big enough.

Since zis an essential point of L then ω(0, Dj, U) > ε > 0 for infinitely many j. Therefore,ω(0, Ej, U) > εa for such j, which means that µ(W, L) > εa > 0.□Inner points on leaves play the role close to the role of inner points in domains. Forexample, if D is a domain in C and f is a mapping of U onto D then for the leaf L = {f}the inner points are points in the interior of D. The third part of Lemma 4.1 also showsthat every point of cl L, which does not belong to the supp L, is inner.

In particular, forexamples of Stolzenberg and Wermer all points of clusters lying in U 2 are inner.Corollary 4.4. If z is a point of a perfect leaf L which does not belong to supp L thenthere is a minimal non-trivial leaf Lz.Proof.

We consider the set of all principal subleaves passing through z. By Lemma 4.1every totally ordered subset of this set has a minimal element.

So, by Zorn lemma, thereis a minimal element in this set.□Lemma 4.2. Let f : U →Cn be a holomorphic mapping such that ||f|| < R, f(0) =z0, b = ||z0|| ̸= 0, and the harmonic measure ω(z0, B1, f(U)) = a > 0 for a ballB1 = B(0, r), r < b.Then for every k > 1 such that kr < b there are a num-ber c = c(k, r, R, a, b) > 0, a number m = m(k, r, R, a, b), k > m > 1, and a pointz1 ∈S(0, mr) where ω(z1, B1, f(U)) > c and ω(z1, Cn \ B(0, kr), f(U)) > c.Proof.

Letu(z) = ln ||z||Rln rR.The function u is plurisuperharmonic, positive on B(0, R), and greater than 1 on B1. If1 < m < k and d = ω(0, f −1(B(0, mr), U) thenu(z0) ≥dln mrRln rR

SEQUENCES OF ANALYTIC DISKS21ord ≤ln bRln mrR.Let v(ζ) = ω(ζ, f −1(B1), U) and let γ = f −1(S(0, mr)). If q is the maximum of v on γthen qd ≥a orq ≥ad = aln mrRln bR≥aln krRln bR= 2s.So there is a point ζ1 on γ where v is greater than s. Let z1 = f(ζ1).Let B2 = B(0, kr) and p = ω(ζ1, f −1(Cn \ B2), U).

Thenu(z1) = ln mrRln rR≥s + (1 −s −p)ln krRln rRorln mrR ≤s ln rR + (1 −s) ln krR −p ln krR = (1 −s) ln k + ln rR −p ln krR .Therefore,p ≥(1 −s) ln k −ln mln krR.Ifm = k(1−s)/2thenp ≥(1 −s) ln k2 ln Rkr= t.If we take c = min(s, t) we get the proof of the lemma.□Theorem 4.4. For a perfect non-trivial leaf L = {fj}:(1) the set of inner points of L is dense in cl L;(2) if a set V is open in cl L and z ∈cl L then ω(z, W, L) = 0 for every set W ⋐Vif and only if any non-trivial subleaf Lz does not intersect V .Proof.

1) We assume that ||fj|| < R/2.Since L is non-trivial we can find a pointz1 ∈cl L such that ||z1 −zL|| = b > 0. For a ball B1 = B(z1, r), r > 0, we considerthe ball B2 = B(z1, 2r).

Since the leaf L is perfect we can find a subsequence fjk suchthat ω(zL, B1, fjk(U)) > a > 0. By Lemma 4.2 there are points ζjk ∈U such that

22EVGENY A. POLETSKYfjk(ζjk) = zk ∈S(z1, mr), 1 < m < 2, and ω(zk, Cn \ B2, fjk(U)) > c > 0. From thesequencegk = fjk ◦Gζjkwe can select a subsequence forming a leaf N = {hl}and let M be the perfection of N.Then zM ∈S(z1, mr) and there are points of cl M lying outside B2.

Otherwise all thispoints are non-essential for the sequence hl, so we can cover the compact intersection ofCn \ B2 and cl N by a finite number of balls such that for each of them its harmonicmeasure at zM tends to 0 as l goes to ∞. This means that ω(zM, Cn\B2, fhl(U)) > c > 0tends to 0 and this is the contradiction.

So the leaf M is non-trivial and, since r isarbitrary, we get the proof of the first part of the theorem.□2) If ω(z, W, L) = 0 then for every sequence gk = fjk ◦Gζjk such points fjk(ζjk)converge to z, all points in V are non-essential and, hence, cannot belong to the perfectionof {gk}. So every subleaf Lz does not intersect W and, therefore, V .Conversely, if ω(z, W, L) > 0 for some W ⋐V then there are essential points in W forthe sequence {gk}.

Therefore, there are points of V in the perfection of this sequence.□Corollary 4.5. If a perfect leaf L is minimal at zL and zL does not belong to supp Lthen for every open set V in cl L there is an open set W containing zL such that everyfunction h ∈H(L) equal to 0 on V is equal to 0 on W.Proof.

Suppose that there is a sequence of points zk converging to zL such that h(zk) ̸=0. By Theorem 4.4 we may assume that points zk are inner and, therefore, there areprincipal subleaves Lk = Lzk = {fkj}, which, by the second part of Theorem 4.4, don’tintersect V .

If we make the perfect leaf M from the sequence {fkk}, then zM = zL andM is non-trivial because supp M belongs to supp L and, hence, doesn’t contain zM. Butcl M is a proper subset because it doesn’t contain V which contradicts to the minimalityof L in zL.□Theorem 4.5.

A leaf L is minimal at any point z ∈D if and only if ω(z, V, L) > 0 forevery inner point z ∈cl L and every open set V ⊂cl L.Proof. If ω(z, V, L) = 0 for some open V and some inner point z then, by Theorem 4.4,a non-trivial subleaf Lz doesn’t intersect V and, hence, L is not minimal at z.Conversely, if L is not minimal at z then there is an open set V which doesn’t intersectsome Lz.

By the second part of Theorem 4.4 this implies that ω(z, V, L) = 0.□

SEQUENCES OF ANALYTIC DISKS23Corollary 4.6. If a leaf is minimal then the algebra H(L) is analytic.Proof.

Follows immediately from Theorem 4.5, 4.4 1), and 4.3.□In his paper [6] Goldmann proves, in different terminology, that the leaf correspondingto the Wermer’s example is minimal. So such leaves exist.

Nevertheless, it seems that ingeneral situation minimality does not happen frequently. If we consider the leaf L fromExample 3.1 then the only part of this leaf where we can guarantee that the algebra H(L)is analytic is the disk {z2 = 0}.

To find an approach to this situation let us introducethe new definition. If L = {fj} is a leaf then the midrib of L is the closure of the set ofall points z in cl L such that ω(z, W, L) > 0 for every neighbourhood W of zL.

For theleaf L mentioned above the midrib is the disk {z2 = 0}.Corollary 4.7. If a perfect leaf L is minimal at zL and zL does not belongs to supp Lthen every holomorphic function h on L equal to 0 on an open set V in cl L is equal to0 on the midrib of L.Proof.

By Corollary 4.5 the function h is equal to 0 a neighbourhood W of zL and, bytwo constants theorem, is equal to 0 on the midrib.□I do not know whether midribs are leaves or not, and I also do not know whether theyare always nontrivial. The following theorem, which can be applied to the Wermer’sexample, gives sufficient criterion for the existence of a nontrivial midrib.Theorem 4.6.

Let L = {fj} be a leaf and let h be a function from H(L) such thath(zL) ̸= h(z) when z ̸= zL is in cl L. We suppose that zL does not belong to supp L. IfK is a connected component, containing h(zL), of the complement of supp L in C andh(z) ∈K for some point z ∈cl L then belongs to the midrib of L.Proof. We fix a neighbourhood W of zL and a point z ∈cl L such that h(z) ∈K.

LetW ∗= h(W). Since h(zL) ̸= h(w) for all w ̸= zL we can find a neighbourhood V ⋐Wof h(zL) such that h−1(V ) ⋐W.

Let us take an approximation g of h by functionsholomorphic on cl L such that g(z) belongs to the connected component K′, containingg(zL), of the complement of the set g(supp L) in C and g−1(V ) ⋐W. We consider onK′ the functionu(ζ) = ω(ζ, g(W), K′)which is, evidently, greater than 0 at g(z).

Sinceω(z, W, L) ≥u(g(z)) > 0

24EVGENY A. POLETSKYthe point z belongs to the midrib.□It follows from this theorem that the leaf generated by the example of Wermer [14]coincides with its midrib.§5. Polynomial hullsLet D be a domain in Cn and let ψ be an upper semicontinuous function on D. Weconsider the functionPψ = u(z) = inf 12π2πZ0ψ(f(eiθ)) dθ,where the infimum is taken over all mappings f ∈A(U, D) with f(0) = z.

It was provedin [8] that u(z) is plurisubharmonic andu(z) = sup{v(z) : v ≤ψ and v is psh }.Let E ⊂D be an open set. Then the characteristic function χE of E is lower semicon-tinuous and we define the pluri-harmonic measure ω(z, E, D) of the set E with respectto D to be equal to P(−χE).

If F is an arbitrary set in D then we define the pluri-harmonic measure ω(z, F, D) to be equal to sup ω(z, E, D), where the supremum is takenover all open sets E containing F. This function may be not upper semicontinuous, sowe take its regularizationω∗(z, F, D) = limw→z ω(z, F, D),which is plurisubharmonic.If X is a compact in Cn then we shall denote by ˆX the polynomial hull of X. Itfollows from Theorem 3.3 that if a leaf L is perfect and X = supp L then cl L ⊂ˆX.If X ⊂D and D is a Runge domain then ˆK coincides with the holomorphic envelopeof K in D. The following theorem was proved in [9]. For the reader’s convenience weshall supply the proof.Theorem 5.1.

Let D be a Runge domain and K be a compact in D. Then z0 ∈ˆK ifand only if for every open set E ⊃K the pluri-harmonic measure ω(z0, E, D) = −1.Proof. Let z0 ∈ˆK and E is an open set containing K.Then there is a decreasingsequence of continuous psh functions vk(z) on D converging to ω(z, E, D) pointwise on

SEQUENCES OF ANALYTIC DISKS25D [11]. By [2] for every ε > 0 there are holomorphic functions fjk, j = 1, 2, .

. ., Nk onD and positive numbers ajk such thatmaxj {ajk ln |fjk|} ≤vk ≤maxj {ajk ln |fjk|} + εon K. Since D is a Runge domain we may assume that all fjk are polynomials.

Wecan take k so big that vk ≤−1 + ε on K. Then vk(z0) ≤−1 + 2ε and this means thatω(z0, E, D) = −1.Conversely, let g be a polynomial. For every ε > 0 we consider the open set E ={z ∈D : |g(z)| < supK|g(w)| + ε} ⊃K.

If ω(z0, E, D) = −1 then there is a mappingf ∈A(U, D) with f(0) = z0 such that m{θ : f(eiθ) ∈E} > 2π(1 −ε). Therefore,|g(z0)| ≤12π2πZ0|g(f(eiθ))| dθ ≤supK|g(w)| + Mε,where M = supD|g(w)|.

Hence, z0 ∈ˆK.□The following theorem establishes something like analytic structure in ˆX.Theorem 5.2. If X is a compact in Cn and z ∈ˆX \ X then there is a perfect leafL = {fj} such that:(1) z = zL;(2) supp L ⊂X;(3) cl L ⊂ˆX;(4) for every neighbourhood V of ˆX there is an integer k such that analytic disksfj(U) are in V when j > k.Proof.

By Theorem 5.1 we can find open neighbourhoods Wj of X with the intersectionequal to X and analytic disks fj such that the measure of points ζ ∈S with fj(ζ) ∈Wjis greater than 2π −j−1 and fj(0) = z. Taking, if necessary, a subsequence we mayassume that the sequence L = {fj} is a leaf.

Evidently, supp L ⊂X. We may alsoassume that L is perfect, otherwise we replace L by its perfection.By the construction, z = zL and supp L ⊂X.

Since cl L belongs to the polynomialhull of supp L, we see that cl L ⊂ˆX. Therefore, for every neighbourhood V of ˆX thereis an integer k such that analytic disks fj(U) are in V when j > k.□

26EVGENY A. POLETSKY§6. Maximal functionsThe plurisubharmonic (psh) function u on a domain D ⊂Cn is called maximal if forevery domain G ⋐D and every psh function v on D the inequality v ≤u on ∂G impliesthe inequality v ≤u on G. Bedford and Kalka proved in [1] that if a maximal functionu is in C2(D) then through every point z ∈D we may draw a holomorphic disk suchthat the restriction of u to this disk is harmonic.

The set of all this disks is called theMonge–Amp´ere foliation for u.It was proved in [9, Lemma 2.2.3, Theorem 2.2.1] that maximal functions on stronglypseudoconvex domains can be obtained as solutions of variational problems for mappingsof the unit disk. More precisely:Theorem 6.1.

Let D be a strongly pseudoconvex domain in Cn and let ϕ < M be acontinuous function on ∂D. We let ϕ to be equal to M on D. LetΦ(f) = 12π2πZ0ϕ(f(eiθ)) dθbe a functional on the space A(U, D) of all holomorphic mappings f : U →D.

Then thefunction u(z) = Pϕ = inf Φ(f), where the infimum is taken over all f ∈A(U, D) suchthat f(0) = z, is a maximal psh function on D, continuous up to the boundary, and equalϕ on ∂D.Moreover, for every point z ∈D there is a sequence {fj} of holomorphic mappingsfj : U →D such that fj(0) = z, supp µ(dz, fj) belongs to ∂D, and Φ(fj) →u(z) asj →∞.As the following example shows the existence of Monge–Amp´ere foliations in the senseof Bedford and Kalka, is not valid in the general case.Example 6.1. Let K be a compact in the boundary of the unit ball B, such thatits polynomial hull ˆK ⊂B doesn’t contain analytic disks.

(One can use examples ofStolzenberg and Wermer to build such K.) If ϕ is a continuous negative function onS = ∂B equal to −1 on K and greater than −1 everywhere else, then we can find amaximal function v which is continuous in B and is equal to ϕ on S. By Theorem 5.1the function v = Pϕ = −1 only on ˆK. If z ∈ˆK and f : U →D is a holomorphic mappingsuch that f(0) = z and v ◦f is harmonic then v ◦f ≡−1 and, hence, f(U) ⊂ˆK.

But

SEQUENCES OF ANALYTIC DISKS27ˆK does not contain analytic disks, so f is trivial, which means that the Monge–Amp´erefoliation for v does not cover ˆK.So to extend the result of Bedford and Kalka to the general case we have to look formore general constructions.Let L = {fj} be a leaf such that supp L ⊂∂D and let G ⊂D be another domain, con-taining an inner point z of L. For a principal subleaf Lz = {gj} we consider a connectedcomponent Gj, containing the origin, of the set g−1j (G) and let hj be a holomorphiccovering mapping of Gj by U such that hj(0) = 0. Then the restrictionof L on Gat z is the perfection of the sequence {gj ◦hj}.

We say that an upper semicontinu-ous function u on cl L is maximal if for all domains G and H such that G ⋐H ⋐Dand every plurisubharmonic (resp. plurisuperharmonic) function v on H, the inequalityu ≥v (resp.

u ≤v) on the intersection of cl L and ∂G implies that u(z) ≥v(z) (resp.u(z) ≤v(z)) on the intersection of cl L and G.Theorem 6.2. Let D be a bounded domain in Cn and let u be a continuous plurisub-harmonic function on D. We consider the functionalΦ(f) = 12π2πZ0u(f(eiθ)) dθon the set A(U, D).

Suppose that for a point z0 ∈D there is a leaf M = {fj} such thatfj ∈A(U, D), z0 = zM, supp M ⊂∂D andu(z0) = limj→∞Φ(fj).Then u is maximal on the perfection L of M.Proof. We may assume that L = M. It is clear that u(z) ≤Φ(f) for every z ∈D andfor every f ∈A(U, D) such that f(0) = z.

Let us take arbitrary domains G ⋐H ⋐Dand let v be a plurisubharmonic function on H such that u ≥v on the intersection K ofcl L and ∂G. We may assume that both u and v are less than some number A on H.Suppose that there is a point z in the intersection F of cl L and G such that u(z)

We fix ε > 0 such that u(z) < v(z) −3ε. Since u is continuous we can find aneighbourhood V of K such that u > v −ε on V and since L is perfect we may assumethat if fj(ζ) ∈∂G then fj(ζ) ∈V .

For each j we consider domains Gj = f −1j(G) whichare unions of connected disjoint domains Gjk.

28EVGENY A. POLETSKYTaking averages if necessary we may assume that v is continuous so u(z) < v(z) −εon a ball B = B(z, r). We denote by Bj sets f −1j(B).

The boundary Γjk of domainsGjk consists of points of U and points of S. We denote the first part by γjk, and thesecond by γ′jk. If uj(ζ) = u(fj(ζ)) and vj(ζ) = v(fj(ζ)) then uj > vj −ε on all γjk.Since supp L ⊂∂D harmonic measures aj = ω(0, γ′jk, U) converge to 0.

If Ej is the setof points in Bj where ω(ζ, γ′jk, Gjk) > s > 0 then, evidently,ω(0, Ej, U) < ajs .Let us introduce harmonic functions hj on U with boundary values of uj on S. Sincehj ≥uj on U the functionhj ≥vj −Aajs−εon the set Cj = Bj \ Ej and, therefore,hj > uj −Aajs+ 2ε > uj + εon Cj when j is sufficiently big. Since the leaf L is perfect, harmonic measuresω(0, Bj, U) > a > 0and, therefore,ω(0, Cj, U) > a −ajs > a2when j is sufficiently big.

So if we consider the universal holomorphic covering mappingsgj of U \ Cj by U, and if we take mappings qj = fj ◦gj thenu(z0) ≤Φ(qj) < Φ(fj) −aε2 ,and we get the contradiction.Now if v is plurisuperharmonic and v ≥u on K then, as before, we may assume thatv is continuous on H. Repeating estimates from the first part of the theorem we mayprove that for every r > 0 and for every ε > 0 the harmonic measure of points in Bjwhere vj ≥uj is uniformly greater than 0 in U. For us it is enough to know that thisset is non-empty because the combination of continuity of u and v and arbitrariness ofr and ε implies that v(z) ≥u(z).□The combination of Theorem 6.1 and 6.2 gives us

SEQUENCES OF ANALYTIC DISKS29Theorem 6.3. A continuous plurisubharmonic function u on a strongly pseudoconvexdomain D in Cn is maximal if and only if through every point z ∈D passes the clusterof a perfect leaf L such that:(1) u is maximal on cl L;(2) supp L ⊂∂L.References1.

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Lowater, The Theory of Cluster Sets, Cambridge University Press, 1966.5. N. Danford, J. T. Schwartz, Linear operators, part I: General Theory, Interscience publishers, NewYork, London, 1958.6.

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Poletsky, To the Bloch theorem, Uspekhi Mat. Nauk 41 (1986), 207–208.8.

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52, Part 1, 1991, pp. 163–171.9.

E.A. Poletsky, Holomorphic currents (to appear).10.

W. Rudin, Function theory in the Unit Ball in Cn, Springer-Verlag, Berlin Heidelberg New York,1980.11. A. Sadullaev, Extensions of plurisubharmonic functions from submanifolds, Dokl.

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Mat. 20 (1982), 129-135.Department of Mathematics, 215 Carnegie Hall, Syracuse University, Syracuse, NY13244

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